ln derivative f (x) = ln(x) ⇒ f ' (x) = 1 / x d/dx (a^x) = a^x ln a U substitution ∫du/u = ln | u | + c In integral ∫ ln(x)dx = x ∙ (ln(x) - 1) + c
LET'S DO SOME MATH GUYS! For u substitution, you would take the whole thing before the dx and call that u and then would derive it to get du
and remember if you get to ∫e^u du you know that the integral of e^u is e^u + C and don't forget to plug back in u into the integral.
if du is in the integral solve, but if it isn't you would need some math makeup. solve and then use the Voldemort's line for your limits if it is a definite integral.
so if it was ∫5/2x dx, it would = 5/2 ln x +c. also the position of du and u matters when you integrate it
so if you have f(x)= ln (x+5)/(x-1) you can go either the hard way or the easy way. You can decipher that you can use your properties to put it as ln a - ln b, or use the chain rule which takes forever and can potentially cause errors. you get the same answer as f'(x)= 1/x+5 - 1/x-1, now do this practice quiz...